Aaron Spike acspike

## GaTechKalturaPlaylistDuration.js
// ==UserScript==
// @name          Display total play time in playlist.
// @namespace     gatech.edu
// @version       0.1.0
// @match         *://canvasgatechtest.kaf.kaltura.com/*
// @match         *://cdnapisec.kaltura.com/*
// @grant         none
// @author        Aaron Spike
// @description   1/25/2021
// ==/UserScript==

## GaTechOscarCourseFilter.js
// ==UserScript==
// @name         GaTech Oscar Course Filter
// @namespace    http://tampermonkey.net/
// @version      0.1
// @description  Hide courses that are not on your shortlist in the Oscar Look-up Classes Advanced Search
// @author       aspike3
// @match        https://oscar.gatech.edu/bprod/*
// @icon         https://www.google.com/s2/favicons?domain=gatech.edu
// @grant        none
// ==/UserScript==

## skog.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script type="text/javascript">

Array.prototype.shuffle = function() {
  var i = this.length, j, temp;
  if ( i == 0 ) return this;

## serial_test.py
import select
import serial

serial_dev = '/dev/ttyS1'

select_timeout = 30

port = serial.Serial(serial_dev, timeout=None)
port.flushInput()

## gist:7999179
#target photoshop
main();

function main() {
    if(!documents.length) return;

    var doc = app.activeDocument;

    try{
        var Path = activeDocument.path;

## solution.py
from __future__ import print_function
import random, itertools, sets

## The following represents thought process and has not been subjected to rigorus proof
# Possible pairings can be represented as two varieties of cycles in the permutations of the residents due to the restriction on reflexive pairings.
# For each permutation there is a single cycle through all 7 name and a two disjoint cycles through 3 names and 4 names respectively.
# Total unique pairings are therefore less than 7!*2 = 10080 (less because cycles of a particular permutation generate equivalent pairings)
# There are 7! permutations. Cycles through all 7 that produce equivalent pairings can be elminiated by considering only permutations that begin with a particular digit (eg. 6).
# Therefore unique pairings for cycles through all 7 generates 6! = 720  (ie 7!/7) pairings.
# Considering 3,4 cycles inside of each permutation there should be (7!/(4!*3))(4!/4) = 420

## solution.py
from __future__ import print_function
import random, itertools, sets

## The following represents thought process and has not been subjected to rigorus proof
# Possible pairings can be represented as two varieties of cycles in the permutations of the residents due to the restriction on reflexive pairings.
# For each permutation there is a single cycle through all 7 name and a two disjoint cycles through 3 names and 4 names respectively.
# Total unique pairings are therefore less than 7!*2 = 10080 (less because cycles of a particular permutation generate equivalent pairings)
# There are 7! permutations. Cycles through all 7 that produce equivalent pairings can be elminiated by considering only permutations that begin with a particular digit (eg. 6).
# Therefore unique pairings for cycles through all 7 generates 6! = 720  (ie 7!/7) pairings.
# Considering 3,4 cycles inside of each permutation there should be (7!/(4!*3))(4!/4) = 420

## solution.py
from __future__ import print_function
import random, itertools, sets

## The following represents thought process and has not been subjected to rigorus proof
# Possible pairings can be represented as two varieties of cycles in the permutations of the residents due to the restriction on reflexive pairings.
# For each permutation there is a single cycle through all 7 name and a two disjoint cycles through 3 names and 4 names respectively.
# Total unique pairings are therefore less than 7!*2 = 10080 (less because cycles of a particular permutation generate equivalent pairings)
# There are 7! permutations. Cycles through all 7 that produce equivalent pairings can be elminiated by considering only permutations that begin with a particular digit (eg. 6).
# Therefore unique pairings for cycles through all 7 generates 6! = 720  (ie 7!/7) pairings.
# Considering 3,4 cycles inside of each permutation there should be (7!/(4!*3))(4!/4) = 420

## solution.py
from __future__ import print_function
import random, itertools, sets

## The following represents thought process and has not been subjected to rigorus proof
# Possible pairings can be represented as two varieties of cycles in the permutations of the residents due to the restriction on reflexive pairings.
# For each permutation there is a single cycle through all 7 name and a two disjoint cycles through 3 names and 4 names respectively.
# Total unique pairings are therefore less than 7!*2 = 10080 (less because cycles of a particular permutation generate equivalent pairings)
# There are 7! permutations. Cycles through all 7 that produce equivalent pairings can be elminiated by considering only permutations that begin with a particular digit (eg. 6).
# Therefore unique pairings for cycles through all 7 generates 6! = 720  (ie 7!/7) pairings.
# Considering 3,4 cycles inside of each permutation there should be (7!/(4!*3))(4!/4) = 420

## SpaceBar.pde
// Much code borrowed from the following
// https://github.com/practicalarduino/VirtualUsbKeyboard
// http://playground.arduino.cc/Learning/SoftwareDebounce

#include "UsbKeyboard.h"

#define BUTTON_PIN 12
#define LED_PIN 13
#define KEY_SPACE 0x2C
	// ==UserScript==
	// @name Display total play time in playlist.
	// @namespace gatech.edu
	// @version 0.1.0
	// @match ://canvasgatechtest.kaf.kaltura.com/
	// @match ://cdnapisec.kaltura.com/
	// @grant none
	// @author Aaron Spike
	// @description 1/25/2021
	// ==/UserScript==
	// ==UserScript==
	// @name GaTech Oscar Course Filter
	// @namespace http://tampermonkey.net/
	// @version 0.1
	// @description Hide courses that are not on your shortlist in the Oscar Look-up Classes Advanced Search
	// @author aspike3
	// @match https://oscar.gatech.edu/bprod/*
	// @icon https://www.google.com/s2/favicons?domain=gatech.edu
	// @grant none
	// ==/UserScript==
	<!DOCTYPE html>
	<html lang="en">
	<head>
	<meta charset="UTF-8">
	<script src="http://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
	<script type="text/javascript">

	Array.prototype.shuffle = function() {
	var i = this.length, j, temp;
	if ( i == 0 ) return this;
	import select
	import serial

	serial_dev = '/dev/ttyS1'

	select_timeout = 30

	port = serial.Serial(serial_dev, timeout=None)
	port.flushInput()
	#target photoshop
	main();

	function main() {
	if(!documents.length) return;

	var doc = app.activeDocument;

	try{
	var Path = activeDocument.path;
	from __future__ import print_function
	import random, itertools, sets

	## The following represents thought process and has not been subjected to rigorus proof
	# Possible pairings can be represented as two varieties of cycles in the permutations of the residents due to the restriction on reflexive pairings.
	# For each permutation there is a single cycle through all 7 name and a two disjoint cycles through 3 names and 4 names respectively.
	# Total unique pairings are therefore less than 7!*2 = 10080 (less because cycles of a particular permutation generate equivalent pairings)
	# There are 7! permutations. Cycles through all 7 that produce equivalent pairings can be elminiated by considering only permutations that begin with a particular digit (eg. 6).
	# Therefore unique pairings for cycles through all 7 generates 6! = 720 (ie 7!/7) pairings.
	# Considering 3,4 cycles inside of each permutation there should be (7!/(4!*3))(4!/4) = 420
	// Much code borrowed from the following
	// https://github.com/practicalarduino/VirtualUsbKeyboard
	// http://playground.arduino.cc/Learning/SoftwareDebounce

	#include "UsbKeyboard.h"

	#define BUTTON_PIN 12
	#define LED_PIN 13
	#define KEY_SPACE 0x2C